Ma connection at home is again functioning. I am happy to have solved  
the problem rather quickly.


On 22 Jul 2009, at 13:54, David Nyman wrote:

>
> 2009/7/22 Bruno Marchal <marc...@ulb.ac.be>:
>
>> Ma connection at home is no functioning.
>
> As a linguistic aside, Bruno has cleverly expressed the above
> statement in perfect Glaswegian (i.e. the spoken tongue of Glasgow,
> Scotland - my home town).


Indeed I am following an intense summer school in Glaswegian.

You thought you could make fun of the poor disconnected one?



> Other well-known examples are: "Is'arra
> marra on yer barra Clarra?" (Is that large vegetable on your barrow a
> marrow, Clara?); and "Gie's'a sook on yer soor ploom" (Let me taste
> the "sour plum" (a globular sweet-sour confection) that you are
> presently sucking).
>
> Perhaps he intends to continue further in this vein?


Once, on a list, someone thought I was using slang from New-York! Now  
Glaswegian!

I am afraid I am just writing to quickly, and then when I read myself  
I concentrate so much on the meaning ...
Most of the time I see the spelling errors when I read my mail, never  
when I send it.

Sorry sorry sorry ...

Take care of the sense and the spelling will take care of itself. Well  
*that* does not work!

Bruno



>
>
> David ;-)
>
>> Hi,
>>
>> Ma connection at home is no functioning. So I am temporarily
>> disconnected. I hope I will be able to solve that problem. I am  
>> sending
>> here some little comments from my office.
>>
>> I include some more material for Kim and Marty, and others, just to
>> think about, in case I remain disconnected for some time. Sorry.
>>
>> Bruno
>>
>> Le 22-juil.-09, à 10:27, Torgny Tholerus a écrit :
>>
>>>
>>> Rex Allen skrev:
>>>> Brent,
>>>>
>>>> So my first draft addressed many of the points you made, but it  
>>>> that
>>>> email got too big and sprawling I thought.
>>>>
>>>> So I've focused on what seems to me like the key passage from your
>>>> post.  If you think there was some other point that I should have
>>>> addressed, let me know.
>>>>
>>>> So, key passage:
>>>>
>>>>
>>>>> Do these mathematical objects "really" exist?  I'd say they have
>>>>> logico-mathematical existence, not the same existence as tables  
>>>>> and
>>>>> chairs, or quarks and electrons.
>>>>>
>>>>
>>>> So which kind of existence do you believe is more fundamental?   
>>>> Which
>>>> is primary?  Logico-mathematical existence, or quark existence?  Or
>>>> are they separate but equal kinds of existence?
>>>>
>>>>
>>>
>>> The most general form of existence is: All mathematical possible
>>> universes exist.  Our universe is one of those mathematical possible
>>> existing universes.
>>
>> This is non sense. Proof: see UDA. Or interrupt me when you have an
>> objection in the current explanation. I have explained this many  
>> times,
>> but the notion of universe or mathematical universe just makes no
>> sense. The notion of "our universe" is too far ambiguous for just
>> making even non sense.
>>
>> I could say the same to Brent. First I don't think it makes sense to
>> say that epistemology comes before ontology, given that the ontology,
>> by definition, in concerned with what we agree exist independently of
>> the observer/knower ... Then what you say contradict the results in  
>> the
>> computationalist theory, where the appearances of universe emerges  
>> from
>> the collection of all computations
>>
>> BTW, thanks to Brent for helping Marty.
>>
>> Rex, when you say:
>>
>>> I would say that most people PERCEIVE logico-mathematical objects
>>> differently than they perceive tables and chairs, or quarks and
>>> electrons.  But this doesn't tell us anything about whether these
>>> things really have different kinds of existence.  That we perceive
>>> them differently is just an accident of fate.
>>
>> We perceive them differently because "observation" is a different
>> modality of self-reference than "proving". It has nothing to do with
>> accident or fate. The comp physics is defined by what is invariant,
>> from the "observation" point of view of universal machine. Later this
>> will shown to be given by the 3th, 4th, and 5th hypostases.
>>
>> ==== math lesson ==== (2 posts):
>>
>> Hi,
>>
>> I wrote:
>> <<
>> The cardinal of { } = 0. All singletons have cardinal one. All pairs,
>> or doubletons, have cardinal two.
>>
>> Problem 1 has been solved. They have the same cardinal, or if you
>> prefer, they have the same number of elements. The set of all subsets
>> of a set with n elements has the same number of elements than the set
>> of all strings of length n.
>>
>> Let us write  B_n for the sets of binary strings of length n. So,
>>
>> B_0 = { }
>> B_1 = {0, 1}
>> B_2 = {00, 01, 10, 11}
>> B_3 = {000, 010, 100, 110, 001, 011, 101, 111}
>>
>> We have seen, without counting, that the cardinal of the powerset  
>> of a
>> set with cardinal n is the same as the cardinal of B_n.
>>  >>
>>
>>
>> And now the killing question by the sadistic math teacher:
>>
>> What is the cardinal, that is, the number of element, of B_0, that is
>> the set of strings of length 0.
>>
>> The student: let see, you wrote above B_0 = { },, and you were kind
>> enough to recall that the cardinal of { } is zero (of course, there  
>> is
>> zero element in the empty set). So the cardinal of B_0 is zero.  
>> 'zero"
>> said the student.
>>
>> 'zero' indeed, said the teacher, but it is your note. You are wrong.
>>
>> B_0 is not empty! It *looks* empty, but beware the appearance, it  
>> looks
>> empty because it contains the empty string, which, if you remember  
>> some
>> preceding post is invisible (even under the microscope, telescope,
>> radioscope, ..).
>>
>> A solution could have been to notate the empty string by a symbol  
>> like
>> "_", and write all sequences "0111000100" starting from "_":
>>  _0111000100, with rules __ = _, etc. Then B_0 = { _ },  B_1 = {_0,
>> _1}, etc. But this is too much notation.
>>
>>
>> And now the time has come for contrition when the teacher feels  
>> guilty!
>>
>> Ah...,  I should have written directly something like
>>
>> B_0 = { _  }, with _ representing the empty sequence.
>> B_1 = {0, 1}
>> B_2 = {00, 01, 10, 11}
>> B_3 = {000, 010, 100, 110, 001, 011, 101, 111}
>>
>> OK?
>>
>> Remember we have seen that the cardinal of the powerset of a set  
>> with n
>> elements is equal to the cardinal of B_n, is equal to 2^n.
>>
>> The cardinal of B_0 has to be equal to to 2^0, which is equal to one.
>> Why?
>>
>> if a is a number, usually, a^n is the result of effectuating (a  
>> times a
>> times a time a ... times a), with n occurences of a. For example:  
>> 2^3 =
>> 2x2x2 = 8.
>>
>> so a^n times a^m is equal to a^(n+m)
>>
>> This extends to the rational by defining a^(-n) by 1/a^n. In that  
>> case
>> a^(m-n) = a^m/a^n. In particular a^m/a^m = 1 (x/x = 1 always), and
>> a^m/a^m = a^(m-m) = a^0. So a^0 = 1. So in particular 2^0 = 1.
>>
>> But we will see soon a deeper reason to be encouraged to guess that  
>> a^0
>> = 1, but for this we need to define the product and the  
>> exponentiation
>> of sets. if A is a set, and B is a set: the exponential B^A is a very
>> important object, it is where the functions live.
>>
>> Take it easy, and ask. Verify the statements a^n/a^m = a^(n-m),  
>> with n
>> = 3 and m = 5. What is a*a*a/a*a*a*a*a  "/" = division, and * =  
>> times).
>>
>> Bruno
>>
>>
>>
>> http://iridia.ulb.ac.be/~marchal/
>>
>> -----------------
>>
>> Hi,
>>
>> I am thinking aloud, for the sequel.
>>
>>
>> There will be a need for a geometrical and number theoretical  
>> interlude.
>>
>> Do you know what is a periodic decimal?
>>
>> Do you know that a is periodic decimal if and only if it exists n and
>> m, integers,  such that a = n/m. And that for all n m, n/m is a
>> periodic decimal?
>>
>> Could you find n and m, such that
>> 12.95213213213213213213213213213213213213 ... (= 12.95 213 213 ...)
>>
>> Solution:
>>
>> Let k be a name for 12.95213213213213213213213213213213213213213 ...
>>
>> Let us multiply k by 100 000.
>>
>> 100 000k = 1295213.213213213213213213213213213213213213 ...  =  
>> 1295213
>> + 0.213213213 ...
>>
>> Let us multiply k by 100
>>
>> 100k = 1295.213213213213... = 1295 + 0.213213213213213..
>>
>>
>> We have 100000k - 100k  = 1295213 + 0.213213213... - 1295
>> - 0.213213213... = 1295213 - 1295 = 1293918
>>
>> So 99900k = 1293918
>>
>> Dividing by 99900 the two sides of the egality we get:
>>
>> k = 1293918/99900
>>
>> We have n and m such that k = n/m = 12.95213213213213213...
>> n = 1293918, and m = 99900.
>>
>> This should convince you that all periodic decimal are fractions.
>>
>> Exercice: find two numbers n and m such that n/m =
>> 31,2454545454545454545... = 31, 2 45 45 45 45 ...
>>
>>
>> Convince yourself that for all n and m, n/m gives always a periodic
>> decimal.(hint: when n is divided by m, m bounds the number of  
>> possible
>> remainders).
>>
>> And now geometry (without picture, do them).
>>
>> Do you know that the length of the circle divided by its diameter is
>> PI? (PI = 3.141592...)
>> Do you know that the length of the square divided by its diagonal is
>> the square root of 2? (sqrt(2)= 1,414213562...)
>>    - can you show this?
>>    - can you show this without Pythagorus theorem?  (like in Plato!)
>>
>> Do you know if it exists n and m such that n/m = the square root of 2
>>  (relation with incommensurability)
>> Do you know if the Diophantine equation  x^2 = 2y^2 has a solution?
>>
>> No.
>> I think I will prove this someday, if only to have an example of
>> simple, yet non trivial, proof.
>>
>> This entails that the sqaure root of 2 cannot be equal to any  
>> fraction
>> n/m.
>> And it means the square root of 2 is a non periodic decimal.  (its
>> decimal will provide a good example of a non trivial computable
>> function).
>>
>> Bruno
>>
>> http://iridia.ulb.ac.be/~marchal/
>>
>>>
>>
>
> >

http://iridia.ulb.ac.be/~marchal/




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